Question

An element of occurs in bcc structure with the cell edge 288 pm its density is 7.8 g/cm^3 . calculate the atomic mass of the element ​

Answer 1

Answer:

Volume of unit cell =(288 pm)3=(288×10−10 cm)3=2.389×10−23 cm3

Volume of 208 g of the element =DensityMass​=7.2208​=28.89 cm3

Number of unit cells =Volume of a unit cellTotal Volume​=2.389×10−2328.89​=12.09×1023

For a BCC structure, number of atoms per unit cell =2

∴ Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells

                                                              =2×12.09×1023

                                                              =24.18×1023

                                                              =2.418×1024

Explanation: